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10x^2+x-132=0
a = 10; b = 1; c = -132;
Δ = b2-4ac
Δ = 12-4·10·(-132)
Δ = 5281
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{5281}}{2*10}=\frac{-1-\sqrt{5281}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{5281}}{2*10}=\frac{-1+\sqrt{5281}}{20} $
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